Problem: The equation of a circle $C$ is $x^2+y^2+16x-14y+77 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+16x) + (y^2-14y) = -77$ $(x^2+16x+64) + (y^2-14y+49) = -77 + 64 + 49$ $(x+8)^{2} + (y-7)^{2} = 36 = 6^2$ Thus, $(h, k) = (-8, 7)$ and $r = 6$.